Introduction
The solution of differential equations
Mathematical Modelling and Vector Algebra:
(APPM172)
Mr Mbusi sivenathi.mbusi@nwu.ac.za
Department of Mathematics and Applied Mathematics
North-West University
Week 8: Differential Equations
April 19, 2022
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
prior Knowledge
Continuity
Differentiation
Integration
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Overview
1
Introduction
2
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Objectives and Learning Outcomes
The major objectives of this study unit are:
To introduce the concept of Differential Equations (DE)
To introduce basic techniques for solving DE
At the end of these Lectures, you should be able to solve
first-order differential equations with initial values using methods
of
1
Separation of variables
2
Integrating factors
3
Partial fraction decomposition
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Definition
A differential equation (DE) is an equation which contains an
unknown function and one or more of its derivatives:
y 00 + 2y
= 4x
dp
2t
− 2pt 2 tp = 0
dt
(1)
(2)
The order of a differential equation is determined by the order of
the highest derivative which appears in the equation.
equation (1): second order
equation (2): first order
A function y = y (x) is a solution of a differential equation if the
function and its derivative satisfy the equation.
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
General and Standard Form
The general form of a linear first-order ordinary Differential
Equation (ODE) is
a1 (x)
dy
+ a0 (x)g (y ) = h(x)
dx
(3)
In equation (3)
y is the unknown function
If a1 (x) = 0, it is no longer an differential equation and so
a1 (x) cannot be 0 and
If a0 (x) = 0 it is a variable separated ODE and can easily be
solved by integration, thus in this chapter a1 (x) 6= 0
If h(x) = 0 it is called a homogeneous ODE, and can easily be
solved by separating the variables.
If h(x) 6= 0 it is called a non-homogeneous ODE, and can
easily be solved by the method of integrating factor.
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
A function y is called a solution for a differential equation if the
equation is satisfied when y = y (x) and its derivatives are
substituted into the equation. When solving a differential equation,
we are expected to find all possible solutions for the equation. For
instance if
R
dy
= f (x) then y = f (x)dx = F (x) + c
dx
where F (x) is the anti-derivative of f (x) that is F 0 (x) = f (x) and
c is the integration constant.
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
Introduction
The solution of differential equations
Method of separation of variables
Definition
If h(x) = 0 in equation (3) the resulting differential equation is
separable since it is possible to group the y ’s with dy and x’s with
dx. So
dy
= f (x)g (y )
dx
where f (x) = − aa01 (x)
(x)
Therefore
dy
= f (x)dx =⇒
g (y )
Z
1
dy =
g (y )
Z
f (x)dx
provide that g (y ) 6= 0
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
The file ”separation of variable method” contains Examples for this
subsection.
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
Solve the differential equation
dy
dx
= 3x 2 y 2
dy
√
b
=x y
dx
c
xyy 0 = x 2 + 1
t sec θ
dθ
d
dt =
θe t 2
dp
e
= t 2p − p + t 2 − 1
dt
Solve the initial-value problem
a
1
2
3
y 0 = xe y ,
y (0) = 0
dy
t sin t
=
,
y (0) = −1
dt
y
p
dy
x ln x = y (1 + 3 + y 2 ) ,
dx
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
y (1) = 1
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
Solving Differential Equations with Integrating Factors
If h(x) ≤ 0 and g (y ) = y in equation (3) the resulting differential
equation can be rewriting as follows
dy
+ P(x)y = Q(x)
dx
(4)
h(x)
where P(x) = aa01 (x)
(x) and Q(x) = a1 (x) are continuous functions.
We can solve this differential equation using the technique of an
integrating factor.
I. Factor We multiply both sides of the differential equation (4)
R by the
integrating factor I (x) which is defined as: I (x) = e P(x)dx
G. Sol Multiplying our original differential equation by I(x) we get
that
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
dy
dy
+ P(x)y = Q(x) ⇔ I (x)
+ I (x)P(x)y = I (x)Q(x)
(5
dx
Z
Z dx
dy
+ I (x)P(x)y dx = I (x)Q(x)d
⇔
I (x)
dx
Z
⇔ yI (x) = I (x)Q(x)dx
Z
1
I (x)Q(x)dx
⇔ y=
I (x)
As both
R P(x) and Q(x) in most of the problems you are likely to
meet, I (x)Q(x)dx can usually be found. So the general solution
to the differential equation is found by integrating I (x)Q(x)dx.
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
The file ”Integrating Factor Method” contains Examples for this
subsection.
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
Introduction
The solution of differential equations
Homework 2
Find the solution of the differential equation that satisfies the
given initial condition.
√
1 xy 0 + y =
x,
y (2) = 0
2
3
4
y 0 + y = 1,
y (0) = 1
2
x dy
x
dx − 2y = x ,
2
0
x y + 2xy = ln x,
> 0,
y (1) = 1
y (1) = 2
6
2
t 3 dy
dt + 3t y = cos t,
xy 0 = y + x 2 sin x,
7
(x 2 + 1)y 0 + 3x(y − 1) = 0,
5
Mr Mbusi
y (π) = 0
y (π) = 0
sivenathi.mbusi@nwu.ac.za
y (0) = 2
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
Partial fraction decomposition
Consider the following rational function
P(x)
Q(x)
where both P(x) and Q(x) are polynomials and the degree of P(x)
is smaller than the degree of Q(x)
Partial fractions can only be done if the degree of the numerator is
strictly less than the degree of the denominator. So, once weve
determined that partial fractions can be done, we factor the
denominator as completely as possible. Then for each factor in the
denominator we can use the following table to determine the
term(s) we pick up in the partial fraction decomposition.
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
Introduction
The solution of differential equations
Factor in denominator
ax + b
Term in partial fraction decomposition:
(ax + b)k
A1
ax+b
ax 2 + bx + c
Ax+B
ax 2 +bx+c
(ax 2 + bx + c)k
A1 x+B1
ax 2 +bx+c
Mr Mbusi
P(x)
Q(x)
A
ax+b
+
sivenathi.mbusi@nwu.ac.za
A2
(ax+b)2
+
+ ... +
A2 x+B2
(ax 2 +bx+c)2
Ak
(ax+b)k
+ ... +
Ak x+Bk
(ax 2 +bx+c)k
Mathematical Modelling and Vector Algebra: (APPM172)
=
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
Example
3x+2
decompose the following expression into partial fraction: x(x+1)
Solution
3x+2
A
B
x(x+1) = x + x+1
Multiply through by the common denominator of x(x + 1) gets rid
of all of the denominators
3x + 2 = A(x + 1) + Bx
= Ax + A + Bx
= (A + B)x + A
For the two sides to be equal, the coefficients of the two
polynomials must be equal. So you ”equate the coefficients” to
get:
A+B =3
A=2
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
Thus A = 2 and B = 1, finally
2
1
3x + 2
= +
x(x + 1)
x
x +1
Find the partial fraction decomposition of
Factor the denominator
7x 2 +5x+7
7x 2 +5x+7
= (x−2)(x+1)
2
(x−2)(x 2 +2x+1)
7x 2 +5x+7
(x−2)(x 2 +2x+1)
Solution
7x 2 + 5x + 7
B
C
A
+
+
=
2
(x − 2)(x + 1)
x − 2 x + 1 (x + 1)2
Multiply through by the common denominator of (x − 2)(x + 1)2
gets rid of all of the denominators
7x 2 + 5x + 7 = A(x + 1)2 + B(x − 2)(x + 1) + C (x − 2)
= A(x 2 + 2x + 1) + B(x 2 − x − 2) + C (x − 2)
= (A + B)x 2 + (2A − B + C )x + (A − 2B − 2C )
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
For the two sides to be equal, the coefficients of the two
polynomials must be equal. So you ”equate the coefficients” to
get:
A+B =7
2A − B + C = 5
A − 2B − 2C = 7
Solve the system of equations to get A = 5, B = 2, C = −3
Therefore
5
2
−3
7x 2 + 5x + 7
=
+
+
2
(x − 2)(x + 1)
x − 2 x + 1 (x + 1)2
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)
Introduction
The solution of differential equations
Separable differential equations
Method of Integrating Factors
Partial fraction decomposition
Homework 3
Determine the partial fraction decomposition of each of the
following
8x − 42
a
f (x) = 2
x + 3x − 18
b
h(t) = t 21+t
c
4u 2
(u − 1)(u + 2)2
Solve the differential equation
g (u) =
1
(x 2 + x) dy
dx = −1
2
y 0 (t 2 + 3t − 18) + 2(4t − 21) = 0
Mr Mbusi
sivenathi.mbusi@nwu.ac.za
Mathematical Modelling and Vector Algebra: (APPM172)